I am struggling to show that the following velocity field is axisymmetric: $u=cos(At)\begin{bmatrix}x/2\\y/2\\-z\end{bmatrix}$ where A is constant.
I have attempted converting the field into cylindrical polar co-ordinates to show that there is no $\phi$ dependence however I simply end up with $u = cos(At)\begin{bmatrix}(rcos\phi)/2\\(rsin\phi)/2\\-z\end{bmatrix}$. I cannot see how to remove the $\phi$ dependence but we are told the flow is axisymmetric. Can anyone provide some guiding tips?
Axisymmetric does not mean constant on circles of constant radius, $r$, and height, $z$.
You didn't quite finish converting to cylindrical polar coordinates, $CP$. The expected coordinate layout would be $\begin{bmatrix} r \\ \phi \\ z \end{bmatrix}$. At the position $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$, the $CP$ position is $\begin{bmatrix} \sqrt{x^2 + y^2} \\ \mathrm{atan2}(y,x) \\ z \end{bmatrix} = \begin{bmatrix} r \\ \phi \\ z \end{bmatrix}$. (See this for more on the function atan2.) The flow vector at that point is $\begin{bmatrix} x/2 \\ y/2 \\ -z \end{bmatrix}$, so the $CP$ flow vector is $$ \begin{bmatrix} \sqrt{x^2/4+y^2/4} \\ \mathrm{atan2}(y/2,x/2) \\ -z \end{bmatrix} = \begin{bmatrix} \frac{1}{2}\sqrt{x^2+y^2} \\ \mathrm{atan2}(y,x) \\ -z \end{bmatrix} = \begin{bmatrix} r/2 \\ \phi \\ -z \end{bmatrix} \text{,} $$ so is parallel to the position vector. What does this flow look like? Here's a slice at $At = z = 0$.
Under rotation by the angle $\theta$ around the $z$ axis, the point $\begin{bmatrix} r \\ \phi \\ z \end{bmatrix}$ is taken to the point $\begin{bmatrix} r \\ \phi + \theta \\ z \end{bmatrix}$. The flow vector that is at $\begin{bmatrix} r \\ \phi + \theta \\ z \end{bmatrix}$ prior to the rotation is $\begin{bmatrix} r/2 \\ \phi + \theta \\ z \end{bmatrix}$. The flow vector that is at $\begin{bmatrix} r \\ \phi \\ z \end{bmatrix}$ before the rotation, $\begin{bmatrix} r/2 \\ \phi \\ -z \end{bmatrix}$, is taken by the rotation to $\begin{bmatrix} r/2 \\ \phi +\theta \\ -z \end{bmatrix}$, showing that the flow field is unchanged by rotating the field around the $z$-axis. That is, the radial component of the flow vector is constant on the circles of constant $r$ and $z$ and the direction is radially outward, parallel to the position vector, so rotation of the entire flow field around the $z$-axis leaves the flow field unchanged.